# Difference between revisions of "Distance formula"

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+ | |||

+ | Shortest distance from a point to a line: | ||

+ | the distance | ||

+ | between the line <math>ax+by+c = 0</math> and point <math>(x_1,y_1)</math> is | ||

+ | |||

+ | <math>|ax_1+by_1+c|/\sqrt(a^2+b^2)</math> | ||

+ | |||

+ | Proof: | ||

+ | The equation <math>ax + by + c = 0</math> can be written: | ||

+ | |||

+ | <math>y = -(a/b)x - (c/a)</math> | ||

+ | |||

+ | So the perpendicular line through (x1,y1) is: | ||

+ | |||

+ | <math>x-x_1</math> <math>y-y_1</math> | ||

+ | ---- = ---- = <math>t/\sqrt(a^2+b^2)</math> where t is a parameter. | ||

+ | a b | ||

+ | |||

+ | t will be the distance from the point <math>(x_1,y_1)</math> along the perpendicular line to (x,y). | ||

+ | |||

+ | So | ||

+ | |||

+ | <math>x = x_1 + a \dot t/\sqrt(a^2+b^2)</math> | ||

+ | |||

+ | and | ||

+ | |||

+ | <math>y = y_1 + b \dot t/\sqrt(a^2+b^2)</math> | ||

+ | |||

+ | This meets the given line ax+by+c = 0 where: | ||

+ | |||

+ | a(x1+a.t/sqrt(a^2+b^2)) + b(y1+b.t/sqrt(a^2+b^2)) + c = 0 | ||

+ | |||

+ | ax1 + by1 + c + t(a^2+b^2)/sqrt(a^2+b^2) + c = 0 | ||

+ | |||

+ | ax1 + by1 + c + t.sqrt(a^2+b^2) = 0 | ||

+ | |||

+ | so | ||

+ | |||

+ | t.sqrt(a^2+b^2) = -(ax1+by1+c) | ||

+ | |||

+ | t = -(ax1+by1+c)/sqrt(a^2+b^2) | ||

+ | |||

+ | Therefore the perpendicular distance from (x1,y1) to the line | ||

+ | ax+by+c = 0 is: | ||

+ | |||

+ | ax1 + by1 + c | ||

+ | |t| = ------------- | ||

+ | sqrt(a^2+b^2) |

## Revision as of 11:35, 3 April 2011

The **distance formula** is a direct application of the Pythagorean Theorem in the setting of a Cartesian coordinate system. In the two-dimensional case, it says that the distance between two points and is given by . In the -dimensional case, the distance between and is

*This article is a stub. Help us out by expanding it.*

Shortest distance from a point to a line: the distance between the line and point is

Proof:

The equation can be written:

So the perpendicular line through (x1,y1) is:

---- = ---- = where t is a parameter. a b

t will be the distance from the point along the perpendicular line to (x,y).

So

and

This meets the given line ax+by+c = 0 where:

a(x1+a.t/sqrt(a^2+b^2)) + b(y1+b.t/sqrt(a^2+b^2)) + c = 0

ax1 + by1 + c + t(a^2+b^2)/sqrt(a^2+b^2) + c = 0

ax1 + by1 + c + t.sqrt(a^2+b^2) = 0

so

t.sqrt(a^2+b^2) = -(ax1+by1+c)

t = -(ax1+by1+c)/sqrt(a^2+b^2)

Therefore the perpendicular distance from (x1,y1) to the line ax+by+c = 0 is:

ax1 + by1 + c |t| = ------------- sqrt(a^2+b^2)